Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $r = \dfrac{t^2 + t}{t^2 + 11t + 30} \div \dfrac{t + 1}{-4t - 24} $
Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{t^2 + t}{t^2 + 11t + 30} \times \dfrac{-4t - 24}{t + 1} $ First factor the quadratic. $r = \dfrac{t^2 + t}{(t + 6)(t + 5)} \times \dfrac{-4t - 24}{t + 1} $ Then factor out any other terms. $r = \dfrac{t(t + 1)}{(t + 6)(t + 5)} \times \dfrac{-4(t + 6)}{t + 1} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ t(t + 1) \times -4(t + 6) } { (t + 6)(t + 5) \times (t + 1) } $ $r = \dfrac{ -4t(t + 1)(t + 6)}{ (t + 6)(t + 5)(t + 1)} $ Notice that $(t + 1)$ and $(t + 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ -4t(t + 1)\cancel{(t + 6)}}{ \cancel{(t + 6)}(t + 5)(t + 1)} $ We are dividing by $t + 6$ , so $t + 6 \neq 0$ Therefore, $t \neq -6$ $r = \dfrac{ -4t\cancel{(t + 1)}\cancel{(t + 6)}}{ \cancel{(t + 6)}(t + 5)\cancel{(t + 1)}} $ We are dividing by $t + 1$ , so $t + 1 \neq 0$ Therefore, $t \neq -1$ $r = \dfrac{-4t}{t + 5} ; \space t \neq -6 ; \space t \neq -1 $